To solve this problem we first calculate sin x / x for lim x--> 0,
This can be done by writing sin x in the Taylor series form
sin x = x - x^3/ 3! + x^5/ 5!...
So sin x / x = 1- x^2/ 3! + x^4/ 4!...
Now for lim x ---> 0 all the terms with x become 0 and we are left with 1.
So sin x/ x = 1 for the limit x---> 0.
Therefore for lim x-->0, as sin x / x = 1, it implies sin x = x.
Yes, it is possible for x = 0. But let's prove it!
We'll have to solve the equation x - sin x = 0.
The equation is a transcedental one, so we'll have to differentiate the function f.
Before differentiating, we'll check if the function is continuous. Because f(x) is formed by elementary functions as the linear one, x , and the trigonometric function, sin x, f(x) is a continuous function.
We'll differentiate f(x).
df/dx = 1-cosx
We notice that f(x) is a monotone increasing function.
( -1<cosx<1), so the difference 1 -cos x>0 =>f(x)>0, so f(x) is an injection.
We can also do a very simple calculus:
Because f(x) is an one-to-one, x=0 is the only solution for the equation x-sinx=0.
So, yes, x = sin x, for x = 0.
No, sinx < less than x for allx.
But sinx/x = 0 asx --> 0.
Suppose we draw a circle with ce ntre as origin O and rardius r. . Le t P be any point and LeX'OX be the x axis cutting the circle at X' on left and X on right.
Draw a perpenicular from P to A.
Let x = angle XOP.
AP =OP sinx < arch XP = OP x.
Divide by OP both sides:
Therefore sinx < x
Then by definiton